2021 RI Promo Sect A Soln

Raffles Institution Year 5-6 Physics Department   1   2021 Year 5 H2 Physics Promotional Examination Solution Section A Qn Ans Solution 1 A 6 6 FFr v rv    2 11 1 base units of kg m s kg m s(base units of )(base units of ) (m)(m s ) F rv     2 B   For : 0.1 0.1 percentage uncertainty 100% 0.20%( ) (51.0 49.0) XY XY XY         For : 0.1 0.1 percentage uncertainty 100% 10%( ) (51.0 49.0) XY XY XY       For Y and : 0.1 0.1 percentage uncertainty 100% 100% 0.40% 51.0 49.0 XX Y XY XY         3 B Since the ball is thrown and caught at the same height above the ground, 0ys  Taking upwards as positive: 210 2 1sin sin ------- (1)22 yut g t gtug t u    Since Jill is running towards the ball at speed 18.9 m sv  ,  cos cos ------- (2)x dv tsu t d v t ut     where 15 md   22(1) (9.81)(0.49)2: t a n 6 . 3 2(2) 2( ) 2 15 (8.9)(0.49) gt gtu dv t dv t ut       From (1): 1(9.81)(0.49) 21.8 22 m s2sin 2sin6.32 gtu     4 A       2 Consider the forces on the crate and present as a system, 30 2.0 0.30 9.81 3.2335 m s2.0 0.30 TM m gM m a a         Consider forces on the present, 0.30 3.2335 9.81 3.91 3.9 N Nm gm a Nm ag     

Raffles Institution Year 5-6 Physics Department   2   5 B By the conservation of energy before the projectile explodes, decrease in K.E. = increase in G.P.E.       2 2 1 1122 0 2 2 9 . 022 speed at the peak, 14.947 m s (horizontally to the right) mm v m g v    By the conservation of momentum, momentum before explosion = momentum after explosion      X 11 X 1 2 14.947 40 10.106 m s 10 m s speed of X 10 m s mm m v v         (Conservation of momentum can be applied th ough there is an external gravitational force acting on the projectile. This is because the force of explosion between X and Y is much larger than the gravitational force. Thus, the effect on their momentum by the gravitational force is negligible.) 6 C Let the angle that the beam makes with the horizontal be  .        Taking moments about P, cos 9.0 4.0 2.0cos 9.0 4.0 2.0 ----- (1) Mg x k Mgx k              Taking moments about Q, 2.0 cos 6.5 4.0 2.0cos 2.0 6.5 4.0 2.0 ----- (2) Mg x k Mg x k          (2) 2.0 2.5 1: (1) 5.0 2 4.0 2 4.0 1.33 1.3 m3 x x xx x        Alternative solution: 6.5 4.0 1 9.0 4.0 2 P Q T T   Taking moments about C.G., 2.0 22.0 4.0 1.33 1.3 m3 PQ Q P Tx T x Tx xT x      

Raffles Institution Year 5-6 Physics Department   3   7 D Since the block is in equilibrium, resultant force and resultant moment about any point must be zero. For the block not to slide down, there must be frictional force acting upwards along the slope on the block to counter the component of its own weight downwards